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Furkat [3]
3 years ago
11

Find a compact form for generating function of the sequence 4, 4, 4, 4, 1, 0, 1, 0, 1, 0, 1, 0,

Mathematics
1 answer:
Zanzabum3 years ago
5 0

The generating function for this sequence is

f(x)=4+4x+4x^2+4x^3+x^4+x^6+x^8+\cdots

assuming the sequence itself is {4, 4, 4, 4, 1, 0, 1, 0, ...} and the 1-0 pattern repeats forever (as opposes to, say four 4s appearing after every four 1-0 pairs). We can make this simpler by "displacing" the odd-degree terms and considering instead the generating function,

f(x)=3+4x+3x^2+4x^3+\underbrace{(1+x^2+x^4+x^6+x^8+\cdots)}_{g(x)}

where the coefficients of g(x) follow a much more obvious pattern of alternating 1s and 0s. Let

g(x)=\displaystyle\sum_{n=0}^\infty a_nx^n

where a_n is recursively given by

\begin{cases}a_0=1\\a_1=0\\a_{n+2}=a_n&\text{for }n\ge0\end{cases}

and explicitly by

a_n=\dfrac{1+(-1)^n}2

so that

g(x)=\displaystyle\sum_{n=0}^\infty\frac{1+(-1)^n}2x^n

and so

\boxed{f(x)=3+4x+3x^2+4x^3+\displaystyle\sum_{n=0}^\infty\frac{1+(-1)^n}2x^n}

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(A-B)n(A-C) ={ }

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2 years ago
Which of the following statements about the volume of a sphere and the volume of a cone are true? Select two answers.
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3 years ago
Suppose a basketball player has made 231 out of 361 free throws. If the player makes the next 2 free throws, I will pay you $31.
statuscvo [17]

Answer:

The expected value of the proposition is of -0.29.

Step-by-step explanation:

For each free throw, there are only two possible outcomes. Either the player will make it, or he will miss it. The probability of a player making a free throw is independent of any other free throw, which means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

Suppose a basketball player has made 231 out of 361 free throws.

This means that p = \frac{231}{361} = 0.6399

Probability of the player making the next 2 free throws:

This is P(X = 2) when n = 2. So

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{2,2}.(0.6399)^{2}.(0.3601)^{0} = 0.4095

Find the expected value of the proposition:

0.4095 probability of you paying $31(losing $31), which is when the player makes the next 2 free throws.

1 - 0.4059 = 0.5905 probability of you being paid $21(earning $21), which is when the player does not make the next 2 free throws. So

E = -31*0.4095 + 21*0.5905 = -0.29

The expected value of the proposition is of -0.29.

3 0
2 years ago
A store has clearance items that have been marked down by 60% they having a sale, advertising an additional 30% off clearance it
bagirrra123 [75]

The original price is 100% of the price. If the price is marked 60% off, then you pay 40% of the original price.

An item costs x dollars.

With the 60% off discount, it now costs 40% of x, or 0.4x.

Now you apply a 30% discount.

For the second discount, consider the price 0.4x to be the new original price. If the price is now discounted 30%, you will pay 70% of the new original price.

Start with 0.4x.

Now calculate 70% of 0.4x.

70% of 0.4x = 0.70 * 0.4x = 0.28x

After applying the 60% discount and the 30% discount, the item that originally cost x now costs 0.28x. 0.28x is the same as 28% of x. The amount you pay is 28% of the original price.

Answer: 28%

3 0
3 years ago
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Amanda [17]
Adding more short sentences.
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